Thermal Energy created by block sliding down ramp

  • #1
1. A 1300 N crate slides 15 m down a ramp that makes an angle of 40 degrees with the horizontal.


2. F therm = Fk[tex]\Delta[/tex][tex]X[/tex]


3. I understand how to solve this with the coefficient of friction given, but how can I without the problem stating [tex]\mu[/tex]?
 

Answers and Replies

  • #2
vela
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Without more information, you can't.
 
  • #3
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You can solve this because to get a constant velocity down the ramp, there must be 0 net acceleration. You can solve for the acceleration down the ramp due to gravity but doing gsin([tex]\theta[/tex]),or 9.8m/s2*sin(40[tex]\circ[/tex]), which gives you 6.3m/s2. Since this is the acceleration downwards, friction must produce the same acceleration up the ramp to produce the constant velocity. Simply multiply this acceleration by the mass of the object (1300N/9.8m/s2=132.653), then by 15 meters.
You get about 12534.358 J, or 1.3*104.
 
  • #4
vela
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Who said anything about the crate moving at constant velocity?
 

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